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Q.

A thin rod AB of mass 1 Kg and length 2m can rotate about a spindle passing through its mid point C and perpendicular to its length. Now a 2 N-m torque is applied to the rod for 10 seconds. Then angular momentum of the rod at the end of 10 S is

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a

20 N-m-s

b

10 N-m-s

c

5 N-m-s

d

40 N-m-s

answer is A.

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Detailed Solution

dLdt=τ, ∫0LdL=τ∫0tdt⇒L=τt=2×10 N−m−s=20 N−m−s
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